Trigonometric Roots and Substitutions: Solving Equations in Terms of Pi

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SUMMARY

The discussion centers on solving the cubic equation 2x^3 - 5x^2 - 4x + 3 = 0, with roots identified as x = -1, 3, and 1/2. The second part involves substituting x = cos(t) into the equation, leading to 2cos^3(t) - 5cos^2(t) - 4cos(t) + 3 = 0 for 0 < t < 2π. The solutions for cos(t) are derived as cos(t) = -1 and cos(t) = 1/2, resulting in an infinite number of real answers. The discussion also touches on the possibility of complex solutions for cos(t) = 3.

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  • Understanding of cubic equations and their roots
  • Knowledge of trigonometric functions, specifically cosine
  • Familiarity with radians and the unit circle
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the properties of cubic equations and their graphical representations
  • Learn about trigonometric identities and their applications in solving equations
  • Explore the concept of complex numbers and their relevance in trigonometric equations
  • Investigate the unit circle and its role in determining angles in radians
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Gughanath
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The first part of this question was to find the roots of the equation 2x^3-5x^2-4x+3=0 i got x=-1, 3, and 1.

but then the second part completely confused me
b) Hence, by substituting x=cost solve the equation 2cos^3t-5cos^2t -4cost+3=0 for 0<t<2"pie", giving your answer in radians in terms of pie.

PLEASE HELP! :confused: :confused: :confused:
 
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Can you solve cos(t)=-1?

Oh, you've solve the cubic equation wrong, 1 of the solutions for x does not work.
 
Last edited:
Gughanath said:
giving your answer in radians in terms of pie.

PLEASE HELP!

The roots are : x=-1,3,1/2.

Just solve : cos(t)=-1 and cos(t)=1/2...this gives of course an infinity of real answers...

If you allow complex numbers, then you can solve cos(t)=3...

NB : You're great, with your help I know how the english pronounciation of [tex]\pi[/tex] double explain it's origin :biggrin:
 

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